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The purpose of DC load line

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saviourm

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Dear ALL

I am studying DC load line of a transistor with a collector resistor of 3K and supply of 15V.
The author of the book calculated the collector current about 5mA. Then he drew a straight line between 5mA on the y-axis and 15V on the x-axis( collector curves Ic vs Vce).The author says that the transistor is operated between these extremities saturation and cutoff.
I understood that when varying the base current of the circuit the output values of collector current(Ic) and collector emitter (Vce) voltage lies on the straight line drawn (provided that the collector resistor and the Vcc remain constant), Am I right?
Please can someone give a reply in simple words what is the purpose of the DC loadline and the said statement if am I rightt?
Many Thanks
Saviour Muscat
 

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With base open (Ib = 0) Vce = Vcc and when Ib is not 0 then Ic flows and max value of Ic for your circuit is Vcc/Rc = 15V/3k = 5 mA. So, Ic can never exceed 5 mA in your circuit. Active region is used for amplifier action as CB junction is reverse biased in active region. In active region Ic increases very very less. In the saturation region (for example 0 to 0.3V for 2N2222A) with Ib not equal to 0 both BE and BC junctions are forward biased. So, load line is drawn between cut-off and saturation value of Ic. When it is mentioned saturation it is peak value of Ic and Ic will be stable after that value and CB junction gets reverse biased. So, on the load line you can see active regions for different Ib values. Depending upon Ib you can have many Q-points or operating points for your amplifier.

The purpose is to select a operating point depending upon the base current range. If you input a ac signal which varies Ib between 10 UA and 40 uA then Ic should vary accordingly on the characteristic curves but in the active region and the point should not lie in the saturation region or cut-off region.
 

I understood that when varying the base current of the circuit the output values of collector current(Ic) and collector emitter (Vce) voltage lies on the straight line drawn (provided that the collector resistor and the Vcc remain constant), Am I right?

Dear Jayanth
thanks for your reply
My quoted statement is it correct or not?
Thanks
Saviourm
 

Dear Jayanth
thanks for your reply
My quoted statement is it correct or not?
Thanks
Saviourm

Hello

Just a reminder, kindly answer my question.

Thanks
SM
 

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